4.1 Exactly one event

1. $$P(atleast\;two\;occurs) = P({E_1} \cap {E_2}) + P({E_2} \cap {E_3}) + P({E_1} \cap {E_3}) - 2P({E_1} \cap {E_2} \cap {E_3})$$

2. $$P(exactly\;two\;occurs) = P({E_1} \cap {E_2}) + P({E_2} \cap {E_3}) + P({E_1} \cap {E_3}) - 3P({E_1} \cap {E_2} \cap {E_3})$$

3. $$P(exactly\;one\;occurs) = P({E_1}) + P({E_2}) + P({E_3}) - 2P({E_1} \cap {E_2}) - 2P({E_3} \cap {E_2}) - 2P({E_1} \cap {E_3}) + 3P({E_1} \cap {E_2} \cap {E_3})$$

A skill is an ability to perform an activity in a competent manner. Skills can be classified into three types, namely functional (transferable), personal traits (attitudes) and knowledge-based. A company calls for an interview. Probability that a person with functional skill gets selected is $0.38$, with nice attitude is $0.26$ and with a good knowledge is $0.14$. The probability of a person with functional skill and good attitude gets selected is $0.12$, with good knowledge and functional skill is $0.03$ and with good knowledge and attitude is $0.09$. As there are very few with all three skills, the probability that a person with all three skills gets selected is $0.01$. Answer the following ($17$ to $20$):

Illustration 17. The panel decides to select a person with exactly one of the skills. Find the probability of selecting under this condition.

Solution:

Let the events be,

$A$ - Person with functional skill gets selected.

$B$ - Person with good attitude gets selected.

$C$ - Person with good knowledge gets selected.

Given,

$$P(A) = 0.38$$

$$P(B) = 0.26$$

$$P(C) = 0.14$$

$$P(A \cap B) = 0.12$$

$$P(A \cap C) = 0.03$$

$$P(B \cap C) = 0.09$$

$$P(A \cap B \cap C) = 0.01$$

$$\begin{equation} \begin{aligned} P(exactly\;one) = P(A) + P(B) + P(C) - 2P(A \cap B) - 2P(A \cap C) - 2P(B \cap C) + 3P(A \cap B \cap C) \\ P(exactly\;one) = 0.38 + 0.26 + 0.14 - 2[0.12] - 2[0.03] - 2[0.09] + 3[0.01] \\ P(exactly\;one) = 0.38 + 0.26 + 0.14 - 0.24 - 0.06 - 0.18 + 0.03 = 0.33 \\\end{aligned} \end{equation} $$

Thus probability of a person with exactly one skill getting selected is $0.33$

Illustration 18. The management feels that an employee must have at least two of these skills for the betterment of management. What is the probability under the above condition?

Solution: We know that,

$$P(atleast\;two) = P(atleast\;one) - P(exactly\;one)$$

Thus,

$$\begin{equation} \begin{aligned} P(atleast\;two) = P(A \cap B) + P(A \cap C) + P(B \cap C) - 2P(A \cap B \cap C) \\ P(atleast\;two) = 0.12 + 0.03 + 0.09 - 2[0.01] = 0.24 - 0.02 \\ P(atleast\;two) = 0.22 \\\end{aligned} \end{equation} $$

Thus the probability that the panel selects a person with at least two skills is $0.22$.

Illustration 19. The Human Resource after a year-long research finds that a person holding exactly two of the above mentioned skills performs better than other employees. What is the probability of selection in this case.

Solution: We know that,

$$P(exactly\;two) = P(atleast\;two) - P(all\;three)$$

$$\begin{equation} \begin{aligned} P(exactly\;two) = P(A \cap B) + P(A \cap C) + P(B \cap C) - 2P(A \cap B \cap C) - P(A \cap B \cap C) \\ = P(A \cap B) + P(A \cap C) + P(B \cap C) - 3P(A \cap B \cap C) \\ = 0.12 + 0.03 + 0.09 - 3[0.01] = 0.24 - 0.03 = 0.21 \\\end{aligned} \end{equation} $$

Thus, the probability of selecting an employee with exactly two skills is $0.21$.

Illustration 20. What is the probability that a person with none of the above mentioned skills gets selected?

Solution: Probability of selecting a person with none of these skills is = $1$ - probability of at least one.

Probability of at least one = $P(A \cup B \cup C)$

$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$

$$\begin{equation} \begin{aligned} P(A \cup B \cup C) = 0.38 + 0.26 + 0.14 - 0.12 - 0.03 - 0.09 + 0.01 \\ P(A \cup B \cup C) = 0.55 \\ P(\bar A \cap \bar B \cap \bar C) = 1 - P(A \cup B \cup C) \\ P(\bar A \cap \bar B \cap \bar C) = 1 - 0.55 = 0.45 \\\end{aligned} \end{equation} $$

Thus, the probability that a company selects an employee with none of the above skills is, $0.45$.

Question 9. If $A$, $B$ and $C$ are three events, such that probability of occurrence of at least one event, $A$ or $B$ is $1 - p$, that of $B$ or $C$ is $1 - p$ and that of $A$ or $C$ is $1 - 2p$. The probability of all three occurring simultaneously is ${p^2}$, then find the range of probability for occurrence of at least one event.

Solution: Given,

$$\begin{equation} \begin{aligned} P(exactly\;one\;out\;of\;A\;or\;B) = P(A) + P(B) - 2P(A \cap B) = 1 - p\;\;\; \to (i) \\ P(exactly\;one\;out\;of\;B\;or\;C) = P(B) + P(C) - 2P(B \cap C) = 1 - p\;\;\; \to (ii) \\ P(exactly\;one\;out\;of\;A\;or\;C) = P(A) + P(C) - 2P(A \cap C) = 1 - 2p \to (iii) \\ P(A \cap B \cap C) = {p^2}\; \to (iv) \\\end{aligned} \end{equation} $$

Adding (i), (ii) and (iii),

$$2\left[ {P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C)} \right] = 3 - 4p$$

Probability that at least one out of the three occurs be $m$,

$$\begin{equation} \begin{aligned} m = P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) \\ m = {{3 - 4p} \over 2} + {p^2} = {p^2} - 2p + {3 \over 2} = {(p - 1)^2} + {1 \over 2} > {1 \over 2} \\\end{aligned} \end{equation} $$

By definition of probability, $$0 \le P(A) \le 1$$

Thus,

$${1 \over 2} < m \le 1$$

Question 10. If there are three events $A$, $B$ and $C$, probability of occurrence of at least one event, $A$ or $B$ is same as the probability of occurrence of at least one event, $C$ or $B$ and probability of occurrence of at least one event, $A$ or $C$. If each of these is $x$ , and probability of occurrence of all three simultaneously is ${x^2}$, where $0 < x < {1 \over 2}$, then find probability of occurrence of at least one of the three events $A$, $B$ and $C$.

Solution: Given,

$$\begin{equation} \begin{aligned} P(exactly\;one\;out\;of\;A\;or\;B) = P(A) + P(B) - 2P(A \cap B) = x\; \to (i) \\ P(exactly\;one\;out\;of\;B\;or\;C) = P(B) + P(C) - 2P(B \cap C) = x\; \to (ii) \\ P(exactly\;one\;out\;of\;A\;or\;C) = P(A) + P(C) - 2P(A \cap C) = x \to (iii) \\ P(A \cap B \cap C) = {x^2}\;\;\; \to (iv)\; \\\end{aligned} \end{equation} $$

Adding (i), (ii) and (iii),

$$2\left[ {P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C)} \right] = 3x$$

Probability that at least one out of the three occurs be $p$,

$$p = P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$

$$P(A \cup B \cup C) = {{3x + 2{x^2}} \over 2}$$

Question 11. The probabilities that a student passes in Science, Social and Mathematics be $a$, $b$, $c$ respectively. The student has $50\% $ chance of passing in at least two, $40\% $ chance of passing in exactly two and $75\% $ chances of passing in at least one. Find the value of $a + b + c$ and also find the chance of passing in exactly one.

Solution: Let $A$, $B$ and $C$ the events of a student passing in Science, Social and Mathematics respectively.

Given that,

$P(A) = a$

$P(B) = b$

$P(C) = c$

$P(A \cup B \cup C) = 0.75$

$$\begin{equation} \begin{aligned} P(atleast\;two) = P(A \cap B) + P(B \cap C) + P(A \cap C) - 2P(A \cap B \cap C) = 0.5 \\ P(exactly\;two) = P(A \cap B) + P(B \cap C) + P(A \cap C) - 3P(A \cap B \cap C) = 0.4 \\ P(atleast\;two) - P(exactly\;two) = P(A \cap B \cap C) = 0.5 - 0.4 = 0.1 \\\end{aligned} \end{equation} $$

Thus,

$$P(A \cap B) + P(B \cap C) + P(A \cap C) - 2P(A \cap B \cap C) = P(A \cap B) + P(B \cap C) + P(A \cap C) - 2(0.1) = 0.5$$

$$ \Rightarrow P(A \cap B) + P(B \cap C) + P(A \cap C) = 0.7$$

Now,

$$\begin{equation} \begin{aligned} P(A \cup B \cup C) = 0.75 \\ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) \\ P(A) + P(B) + P(C) - [P(A \cap B) + P(B \cap C) + P(A \cap C)] + P(A \cap B \cap C) = P(A \cup B \cup C) \\ P(A) + P(B) + P(C) - 0.7 + 0.1 = 0.75 \\ P(A) + P(B) + P(C) = 1.35 \\ a + b + c = 1.35 \\ a + b + c = {{27} \over {20}} \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} P(exactly\;one) = P(A) + P(B) + P(C) - 2P(A \cap B) - 2P(B \cap C) - 2P(A \cap C) + 3P(A \cap B \cap C) \\ P(exactly\;one) = a + b + c - 2[P(A \cap B) + P(B \cap C) + P(A \cap C)] + 3P(A \cap B \cap C) \\ P(exactly\;one) = 1.35 - 2[0.7] + 3(0.1) \\ P(exactly\;one) = 1.35 - 1.4 + 0.3 \\ P(exactly\;one) = 0.25 = 25\% \\\end{aligned} \end{equation} $$